Question

Three point charges lie in the xy-plane: with q1= 86 μC at origin (0,0), q2= 32μC at the point (2.5m, 0), and q3=−53μC at the point (1.5m, 2.2m). Find the net force on q1?

Answer 1

`F_(net) = -0.7 \hat i + 4.77 \hat j`

Step-by-step explanation:

Force due to q2 on q1 is along - X direction due to repulsion between them

so we have

`F_(12) = (kq_1q_2)/(r^2)`

`F_(12) = ((9* 10^9)(86 \mu C)(32 \mu C))/(2.5^2)`

`F_(12) = 3.96 N (-\hat i)`

Now force between q1 and q3 is given as

`F_(13) = (kq_1q_3)/(r^2) ((1.5\hat i + 2.2 \hat j))/(√(1.5^2 + 2.2^2))`

`F_(13) = ((9* 10^9)(86 \mu C)(53 \mu C))/((1.5^2 + 2.2^2) ((1.5\hat i + 2.2 \hat j))/(√(1.5^2 + 2.2^2))`

`F_(13) = (5.78)((1.5\hat i + 2.2 \hat j))/(2.66)`

`F_(13) = (2.17)(1.5\hat i + 2.2 \hat j)`

Now net force on q1 is given as

`F_(net) = F_(12) + F_(13)`

`F_(net) = 3.96(-\hat i ) + (3.26 \hat i + 4.77 \hat j)`

`F_(net) = -0.7 \hat i + 4.77 \hat j`